Clever Geek Handbook

Passing command line arguments through Bash

While brushing on bash (it was a while) I was surprised to notice that the execution of this code is being saved as script.sh:

echo "Arg 0 to script.sh: $0"
echo "Arg 1 to script.sh: $1"
function echo_args
{
    echo "Arg 0 to echo_args: $0"
    echo "Arg 1 to echo_args: $1"
}
echo_args

like this:

>> ./script.sh argument

output this:

Arg 0 to script.sh: ./script.sh
Arg 1 to script.sh: argument
Arg 0 to echo_args: ./script.sh
Arg 1 to echo_args:

I was surprised to see $ 0 from script.sh passed as $ 0 to echo_args when $ 1 is not handled similarly. It seems to me that it should be both.

Any clarification is appreciated.

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1 answer


$ 0 is a "special parameter" in bash that always evaluates the script name and is set at the beginning of the script. It looks like it also means that it is somewhat global as it cannot be reassigned.

For more information, this is a pretty good recommendation:



http://bash.cyberciti.biz/guide/$0

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